Let $f(z)=\sqrt[3]{\sqrt{z^2-z^3}+z}$, $g(z)=\sqrt{\frac{f(z)}{z}+\frac{1}{f(z)}}$ , then how to prove that for $\ 0<z<1$:

$$\small \, _4F_3\left(\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{9}{8};\frac{5}{6},\frac{7}{6},\frac{9}{6};z\right)=\frac{4 \sqrt[4]{2} }{3 \sqrt{3}}\left(g(z)-\sqrt{-\frac{f(z)}{z}+\frac{2 \sqrt{2}}{z g(z)}-\frac{1}{f(z)}}\right)^{3/2}$$

This formula is collected from Wolfram Functions site but no proof is offered there. I think it will be worthy to find a rigorous proof, as well as to investigate the motivation of this formula in order to generate similar algebraic closed-forms of generalised hypergeometric series (very likely, their $_2F_1$ counterparts are those in Schwarz's list). Any help will be appreciated!


The following is a somewhat sprawling proof of the formula, but there are really only two main steps. The key "observation" (!) is that the expression in the parentheses on the right hand side is proportional to one of the roots of the quartic polynomial $z x^4 - 4 x + 3$. Thus we can first prove that the expression in the parentheses does indeed solve the quartic, then we prove that the given hypergeometric function equals this particular function of the quartic root.

For the first step, we can simply use the formula for quartic roots. The formulae on Wikipedia are written for a general quartic $a x^4 + b x^3 + c x^2 + d x + e$, and are quite cumbersome to repeat here, but for us $b = c = 0$, so many of those expressions simplify. Leaving some of the intermediate verification to you, I will state that $\Delta_0 = 36z$, $\Delta_1 = 432z$ and $p = 0$, so $$ Q = 6 \sqrt[3]{z + \sqrt{z^2-z^3}} = 6 f(z)\ , $$ which means that $$ S=\frac{1}{\sqrt{2}} \sqrt{\frac{1}{f(z)} + \frac{f(z)}{z}}\ . $$ Plugging this and $q = - 4/z$ into the final formula for the roots, we get $$ x_{u, v} = \frac{1}{\sqrt{2}}\left(u\sqrt{\frac{1}{f(z)} + \frac{f(z)}{z}}+v\sqrt{-\left(\frac{1}{f(z)} + \frac{f(z)}{z}\right)+2\sqrt{2} v\left/\sqrt{\frac{1}{f(z)} + \frac{f(z)}{z}}\right.}\right)\ . $$ Taking $u=\pm 1$ and $v = \pm 1$ gives us the four roots. The root that appears in your expression is $x_{1,-1}$.

(The far-fetched "observation" at the beginning of the answer requires knowing the quartic formula beforehand. Since the quartic formula contains $$ Q = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}\ , $$ one might surmise that $f(z)\propto Q$ in the root formula for a quartic because the powers and roots match up. For that to happen, we need both $\Delta_0$ and $\Delta_1$ to be equal to $z$. In order to further match up the form of $S$, we also need $p = (8ac - 3 b^2)/8a^2 = 0$. To satisfy the latter constraint we make the guess that $b =c = 0$. The former constraint then forces us to guess that $a$ is proportional to $z$.)

Next we would like to show that ${}_4 F_3\left(\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{9}{8};\frac{5}{6},\frac{7}{6},\frac{9}{6};z\right) = \left(\frac{4}{3} x_{1,-1}(z)\right)^{3/2}$. (Let's denote this expression by $(\star)$ for later references.) To do so, we first prove that the function on the right hand side satisfies a generalized hypergeometric differential equation , and then find certain extra initial conditions which will gives us a particular solution that equals the left hand side.

The particular case of generalized hypergeometric equations that we should look at is $$ \begin{multline} z \frac{d}{dz} \left(z \frac{d}{dz} + b_1 - 1\right) \left(z \frac{d}{dz} + b_2 - 1\right) \left(z \frac{d}{dz} + b_3 - 1\right) y(z)\\ = z \left(z \frac{d}{dz} + a_1\right) \left(z \frac{d}{dz} + a_2\right) \left(z \frac{d}{dz} + a_3\right) \left(z \frac{d}{dz} + a_4\right) y(z)\ , \end{multline} $$ where $a_1 = \frac{3}{8}, a_2=\frac{5}{8}, a_3 = \frac{7}{8}, a_4=\frac{9}{8}$ and $b_1 = \frac{5}{6},b_2 = \frac{7}{6},b_3 = \frac{9}{6}$. It has $4$ linearly independent solutions, one of which is ${}_4 F_3(a_1, a_2, a_3, a_4; b_1, b_2, b_3;z)$. The other solutions are of the form $z^{1-b_i} {}_4 F_3(1+a_1-b_i, 1 + a_2-b_1,1+a_3-b_i,1+a_4-b_i;1+b_1-b_i,\dots, 2-b_i;z)$.

We can verify that the $\left(\frac{4}{3} x_{1,-1}(z)\right)^{3/2}$ solves the differential equation by plugging it in, taking all of the derivatives, and simplifying. However, this is a rather difficult task even with the help of Mathematica. Here I present a different method that goes in the opposite direction, that is, we will build a differential equation called the differential resolvent that $\left(\frac{4}{3} x_{1,-1}(z)\right)^{3/2}$ satisifes, which will turn out to be the generalized hypergeometric equation above.

The construction here is very similiar to the process described in this answer here. Essentially what we will do is write down a linear combination of derivatives of $y(z)$ that is forced to be zero due to certain relations that these derivatives satisify, which are derived from an algebraic equation that $y$ itself satisfies. Since $x(z) = \left(\frac{4}{3} y(z)\right)^{3/2}$ satisfies $z x^4 - 4x +3$, we have the following equation for $y$, $$ \frac{81}{256} z y^{8/3} - 3 y^{2/3} + 3 = 0\ . $$ We can implicitly diffrentiate this equation to express all the derivatives of $y$ in terms of $y$ and $z$. We want to find the coefficients $\mu_i(z)$ that make the following expression zero, $$ \mu_0 y''''(z) + \mu_1 y'''(z) + \mu_2 y''(z) + \mu_3 y'(z) + \mu_4 y(z) + \mu_5\ . $$ Using the expressions of $y^{(n)}(z)$ derived earlier, this can be rewritten as a rational function of $y^{1/3}$ whose numerator is a polynomial of $y^{1/3}$ of some high degree. We can use the algebraic equation for $y$ to reduce the degree of this polynomial down to less than $8/3$. Now we force this expression to be zero, which is to say that the coefficients of each power of $y^{1/3}$ should be $0$. From these conditions we can solve for $\mu_i$ in terms of $z$, and finally obtain the following differential equation. $$ \begin{multline} (z^3-z^4)y''''(z) + \left(\frac{13}{2}z^2-9z^3\right)y'''(z) \\ + \left(\frac{305}{36}z - \frac{615}{32}z^2\right)y''(z) + \left(\frac{35}{24} - \frac{555}{64}z\right)y'(z) - \frac{945}{4096} y(z) = 0\ . \end{multline} $$ We can plug in the parameters of the generalized hypergeometric equation and simplify it to confirm that it is indeed the same as the differential resolvent. This process is rather tedious and not difficult with the aid of Mathematica, so I will not record it here. Nevertheless, to be a bit more explicit, I will say that the equation simplifies to $$ \begin{multline} (z^3-z^4) y''''(z) + [(t_1 + 3)z^2 - (s_1 + 6)z^3]y'''(z) + \\ [(t_1+t_2+1)z-(3s_1+s_2+7)z^2]y''(z) + [t_3 - (s_1+s_2+s_3+1)z]y'(z) - s_4 y(z) = 0 \end{multline} $$ where $s_i$ and $t_i$ are the degree $i$ elementary symmetric polynomials in $a_1,a_2, a_3, a_4$ and $b_1, b_2, b_3$, respectively. Hopefully this is easier to verify by hand.

Finally, to find the particular solution of the differential resolvent that satisifies $(\star)$, we can use the derivatives of the right hand side to provide initial conditions that will pin down the particular solution. These derivatives can be easily evaluated, since at the first step of the construction of the differential resolvent, we have already expressed the derivatives in terms of the values of the function.


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